Does a capacitor's capacitance depend on the charge stored?
Short answer: C depends only on geometry and dielectric (C = ε₀A/d). V doubles instead.
The trap
The wrong idea
“A capacitor's capacitance depends on the charge stored on it.”
It sounds reasonable, and a lot of JEE/NEET aspirants carry it into the exam. Here's a one-question check, then the explanation that makes it click.
Quick check
Try it before you peek
Doubling the charge on a parallel-plate capacitor makes its capacitance:
- Double
- Half
- Unchanged
Show the answer
C. Unchanged
C depends only on geometry and dielectric (C = ε₀A/d). V doubles instead.
The click
Capacitance is defined as C = Q/V, the charge stored per volt of potential difference. Doubling Q does not change this ratio, because V doubles too. For a given capacitor, Q and V are always proportional, and C is the fixed constant of proportionality.
What C depends on is the geometry and the material between the plates. For a parallel-plate capacitor, C = ε₀A/d with vacuum (and very nearly with air) between the plates, and C = Kε₀A/d with a dielectric of dielectric constant K filling the gap. Bigger plates, a smaller gap or a dielectric raise C. Charge does not appear in the formula.
Why V rises with Q: the field between the plates is E = σ/ε₀ = Q/(ε₀A), so more charge means a stronger field, and V = Ed doubles when Q doubles. Their ratio stays the same.
Resistance works the same way: R = V/I, but for a given wire at fixed temperature R depends on the wire itself (R = ρL/A), not on the current. In circuits, remember the two cases: a capacitor connected to a battery keeps V fixed, so changing C changes Q; an isolated capacitor keeps Q fixed, so changing C, for example by inserting a dielectric, changes V. The energy stored is U = ½CV² = Q²/2C.
Got this one? The Knot Scan finds the misconception you don't know you have, in 8 questions.